What are the solutions to the equation k^2+k=12 ?

The solutions to the equation k^2+k=12 are k=3,k=-4

Find the zeros of k^2+k=12

The zeros of k^2+k=12 are k=3,k=-4

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k^{2} + k = 12

k = 3, k = - 4

Solution steps

k^{2} + k = 12

Move

12

to the left side

k^{2} + k - 12 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 12) = 7

k_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- 1 + 7}{2 \cdot 1}, k_{2} = \frac{- 1 - 7}{2 \cdot 1}

k = \frac{- 1 + 7}{2 \cdot 1} : 3

k = \frac{- 1 - 7}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

k = 3, k = - 4

t, 8 \cdot t^{2} - 22 \cdot t + 9 = 0

s^{2} + s - 2 + 1 = 0

x^{2} - 9 x - 16 = 0

10 x^{2} + 37 x + 30 = 0

y^{2} = 6 - 2 y^{2}

What are the solutions to the equation k^2+k=12 ?
The solutions to the equation k^2+k=12 are k=3,k=-4
Find the zeros of k^2+k=12
The zeros of k^2+k=12 are k=3,k=-4